Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies. Express your answer in terms of Ki and Kf.

Answer:

Total time taken(T) = 1.1 hour

Distance = 75.68 km

Explanation:

Given:

Average speed = 68.8 km/h

Constant speed = 98.5 km/h

Rest time = 20 min = 20 / 60 = 0.3333 hour

Find:

Total time taken(T)

Total distance (D)

Computation:

Distance = speed × time

D = 68.8 × t.........Eq1

and

D = 98.5 × [t-0.33]

D = 98.5 t - 32.8333.........Eq2

From Eq1 and Eq2

68.8 t = 98.5 t - 32.83333

29.7 t = 32.83333

t = 1.1

Total time taken(T) = 1.1 hour

Distance = speed × time

Distance = 68.8 × 1.1

Distance = 75.68 km

Answer:

230N/m

Explanation:

Pls see attached file

Answer:

The  induced current is [tex]I = 5.72*10^{-4 } \ A[/tex]

Explanation:

From the question we are told that

     The cross-sectional area is  [tex]A = 8.60 \ cm^2 = \frac{8.60 }{10000} = 8.60 *10^{-4} \ m[/tex]

     The initial value of magnetic field is  [tex]B_1 = 0.500 \ T[/tex]

     The  value of magnetic field  at  time  t     is  [tex]B_f = 2.40 \ T[/tex]

     The number of turns  is  N  =  1  

     The  time taken is   [tex]dt[/tex]=  1.02 \ s  

       The resistance of the loop is  [tex]R = 2.80\ \Omega[/tex]

Generally the induced emf is mathematically represented as

         [tex]e = - \frac{d \phi}{dt }[/tex]

Where  [tex]d \phi[/tex] is the change n the magnetic flux which is mathematically represented as

          [tex]d \phi = N *A * d B[/tex]

Where [tex]dB[/tex] is the change in magnetic field which is mathematically represented as  

          [tex]d B = B_f - B_i[/tex]

substituting values  

         [tex]d B = 2.40 - 0.500[/tex]

         [tex]d B = 1.9 \ T[/tex]

So  

        [tex]d \phi = 1 * 1.9 * 8.60 *10^{-4}[/tex]

       [tex]d \phi = 1.63*10^{-3} \ T[/tex]

So  

      [tex]e = - \frac{1.63 *10^{-3}}{ 1.02 }[/tex]

      [tex]e = - 1.60*10^{-3} \ V[/tex]

     Here the negative only indicates that the emf is acting in opposite direction of the motion producing it so the magnitude of the emf is  

       [tex]e = 1.60*10^{-3} \ V[/tex]

Now the induced current is evaluated as follows

       [tex]I = \frac{e}{R }[/tex]

substituting values  

      [tex]I = \frac{1.60 *10^{-3}}{2.80 }[/tex]

      [tex]I = 5.72*10^{-4 } \ A[/tex]

Answer:

Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox.

Explanation:

The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.

Or

If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.  

On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.

Answer:

The direction of the magnetic field causing this force is

In the plane of the screen and towards the bottom of the egde

Explanation:

This is by applying Fleming s right hand rule which explains that

When a conductor such as a wire attached to a circuit moves through a magnetic field, an electric current is induced in the wire due to Faraday's law of induction. The current in the wire can have two possible directions. Fleming's right-hand rule gives which direction the current flows.

The right hand is held with the thumb, index finger and middle finger mutually perpendicular to each other (at right angles), as shown in the diagram.[1]

The thumb is pointed in the direction of the motion of the conductor relative to the magnetic field.

The first finger is pointed in the direction of the magnetic field. (north to south)

Then the second finger represents the direction of the induced or generated current within the conductor (from the terminal with lower electric potential to the terminal with higher electric potential, as in a voltage source)

Answer:

5.2m/s

Explanation:

Plss see attached file

Answer:

The kinetic energy of the ball 1 is 2.06 J

Explanation:

The kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I is its rotational inertia, ω its angular speed, m its mass and v = its velocity of center of mass.

Let m₁, I₁, v₁, d₁ represent the mass, rotational inertia, speed and diameter of  solid ball 1. and Let m₂, I₂, v₂, d₂ represent the mass, rotational inertia, speed and diameter of  solid ball 2.

Since both objects are spheres, I =2/5mr²

Let r₁ = radius of ball 1 and r₂ = radius of ball 2. Since d₂ = 2d₁

⇒ 2r₂ = 4r₁ ⇒ r₂ = 2r₁

Now, the the kinetic energy of sphere 1 is

K₁ = 1/2I₁ω₁² + 1/2m₁v₁²  ω₁ = v₁/r₁ which is the angular speed of solid ball 1.

K₁ = 1/2(2/5mr²)v₁²/r₁² + 1/2m₁v₁²

K₁ = 1/5m₁v₁² + 1/2m₁v₁²

K₁ = 7/10m₁v₁²

Also, the the kinetic energy of sphere 2 is

K₂ = 1/2I₂ω₂² + 1/2m₂v₂²  ω₂ = v₂/r₂ which is the angular speed of solid ball 2.

K₂ = 1/2(2/5m₂r₂²)v₂²/r₂² + 1/2m₂v₂²

K₂ = 1/5m₂v₂² + 1/2m₂v₂²

K₂ = 7/10m₂v₂²

Now, m₁ = m₂/2 and v₁ = v₂/3

Substituting these into K₁, we have

K₁ = 7/10(m₂/2)(v₂/3)²

K₁ = 7/10 × 1/18m₂v₂²

K₁ = (1/18)(7/10m₂v₂²)

K₁ = K₂/18

K₂ = 37.0 J/18

K₂ = 2.06 J

So, the kinetic energy of the ball 1 is 2.06 J

Answer:

Explanation:

speed of alien spaceship = .1 c

We shall apply formula of relativistic mechanics to solve the problem

relative velocity =

[tex]\frac{v+v_1}{1 -\frac{v\times v }{c^2} }[/tex]

Here v = v₁ = .1 c

relative velocity  = .1c + .1 c / 1 - .1²

= .2 c / .99

= .202 c

The earth would receive the signal at the speed of .202 c .

Answer:

a. Bx is positive

Explanation:

See attached file

Answer:

The magnitude of the gravitational force is 4.53 * 10 ^-7 N

Explanation:

Given that the magnitude of the gravitational force is F = GMm/r²

mass M = 850 kg

mass m = 2.0 kg

distance d = 1.0 m , r = 0.5 m

F = GMm/r²

Gravitational Constant G = 6.67 × 10^-11 Newtons kg-2 m2.

F = (6.67 × 10^-11 * 850 * 2)/0.5²

F = 0.00000045356 N

F = 4.53 * 10 ^-7 N

Answer:

1.24m

Explanation:

See attached file

Complete question:

A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid consists of 2100 turns of wire.

Answer:

The magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.

Explanation:

Given;

length of solenoid, L = 2.4 m

radius of solenoid, R = 1.7 cm = 0.017 m

current in the solenoid, I = 0.19 A

number of turns of the solenoid, N = 2100 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μnI

Where;

μ is permeability of free space = 4π x 10⁻⁷ m/A

n is number of turns per length = N/L

I is current in the solenoid

B = μnI = μ(N/L)I

B = 4π x 10⁻⁷(2100 / 2.4)0.19

B = 4π x 10⁻⁷ (875) 0.19

B = 2.089 x 10⁻⁴ T

Therefore, the magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.

Answer:

Explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y = [tex]\frac{\delta m \lambda d}{a}[/tex] where;

[tex]\delta m[/tex] is the first two diffraction minima = 1

[tex]\lambda[/tex] is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given [tex]\lambda[/tex] = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is  1.643*10⁻⁴cm

Answer:

1.5x10^-1 V

Explanation:

See attached file

Answer:

The magnitude of the induced emf in the coil is 15.3 mV

Explanation:

Given;

number of turns, N = 20 turns

Area of each coil, A = 0.0015 m²

initial magnitude of magnetic field at t₁, B₁ = 4.91 T/s

final magnitude of magnetic field at t₂, B₂ = 5.42 T/s

The magnitude of the induced emf in the coil is given by;

[tex]E = -N\frac{\delta \phi}{\delta t} \\\\E =-N (\frac{\delta B}{\delta t} )A\\\\E = -NA(\frac{B_1-B_2)}{\delta t} \\\\E = NA(\frac{B_2-B_1)}{\delta t} \\\\E = 20(0.0015)(5.42-4.91)\\\\E = 0.0153 \ V\\\\E = 15.3 \ mV[/tex]

Therefore, the magnitude of the induced emf in the coil is 15.3 mV

Answer:

0.44m/s

Explanation:

drift velocity=I/nAq

diameter 12 gauge

wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

V = 0.44m/s

The drift velocity of the electrons should be 0.44 mm/s.

Since the diameter is 0.081 in

So, the  radius = r = 0.081 inch/2

= 0.0405 inch

Now the conversion of inches to cm should be

= 0.0405*2.54

= 0.10287 cm

Now

area = π r2

= 3.14 * (0.10287)2

= 0.0332

Now the velocity should be

v = i/(nqA)

= 20/(8.5e22*1.60e-19*0.0332)

= 0.044 cm/s

= 0.44 mm/s

Learn more about velocity here: https://brainly.com/question/19279241

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  wavelength is  [tex]\lambda = 622 nm[/tex]

Explanation:

  From the question we are told that

    The distance of the slit to the screen is  [tex]D = 5 \ m[/tex]

    The order of the fringe is m  =  6

     The distance between the slit is  [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]

    The fringe distance is  [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]

Generally the for a dark fringe the fringe distance is  mathematically represented as

        [tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]

=>     [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]

substituting values

=>      [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]

=>     [tex]\lambda = 6.22 *10^{-7} \ m[/tex]

       [tex]\lambda = 622 nm[/tex]

Answer:

The  current in the tube is 0.601 A

Explanation:

Given;

diameter of the fluorescent, d = 3 cm

negative charge flowing in the fluorescent tube, -e = 3 x 10¹⁸ electrons/second

positive charge flowing in the fluorescent tube, +e = 0.75 x 10¹⁸ electrons/ second

The current in the fluorescent tube is due to presence of positive and negative charges to create neutrality in the conductor (fluorescent tube).

Q = It

I = Q/t

where;

I is current in Ampere (A)

Q is charge in Coulombs (C)

t is time is seconds (s)

1 e = 1.602 x 10⁻¹⁹ C

3 x 10¹⁸ e/ s = ?

= (3 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

= 0.4806 C/s

negative charge per second (Q/t) = 0.4806 C/s

positive charge per second (Q/t) =  (0.75 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

positive charge per second (Q/t) = 0.12015 C/s

Total charge per second in the tube, Q / t = (0.4806 C/s + 0.12015 C/s)

                                                                I = 0.601 A

Therefore, the  current in the tube is 0.601 A

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

1.6nT [in the negative z direction]

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb[tex]\sqrt{b^2 + L^2}[/tex])                -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 [tex]\sqrt{5.0^2 + 0.02^2}[/tex])

B = (10⁻⁷ x 20 x 0.02) / (5.0 [tex]\sqrt{5.0^2 + 0.02^2}[/tex])

B = (10⁻⁷ x 20 x 0.02) / (5.0 [tex]\sqrt{25.0004}[/tex])

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

B = 1.6nT

Therefore, the magnetic field at the point x = 5.0m  on the x-axis is 1.6nT.

PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.

Answer:

The current will be 18 A

Explanation:

Given;

potential difference, V = 10 V

current between the resistor, I = 2 A

Apply ohm's law;

V = IR

R = V / I

R = 10 / 2

R = 5Ω

Resistance is given as;

[tex]R = \frac{\rho l}{A}[/tex]

where;

ρ is resistivity

l is length

A is area

[tex]R = \frac{\rho l}{A} \\\\R = \frac{\rho l}{\pi r^2} = \frac{\rho l}{\pi (\frac{d}{2}) ^2} = \frac{\rho l}{\pi (\frac{d^2}{4}) }\\\\R = \frac{4*\rho l}{\pi d^2} \\\\R = (\frac{4*\rho l}{\pi } )\frac{1}{d^2} \\\\R = (k)\frac{1}{d^2} \\\\k = Rd^2\\\\R_1d_1^2 = R_2d_2^2\\\\R_2 = \frac{R_1d_1^2}{d_2^2}[/tex]

When the diameter of the resistor is tripled

d₂ = 3d₁

[tex]R_2 = \frac{5*d_1^2}{(3d_1)^2} \\\\R_2 = \frac{5d_1^2}{9d_1^2} \\\\R_2 = 0.556 \ ohms[/tex]

The current is now calculated as;

Apply ohms law;

V = IR

I = V / R

I = 10 / 0.556

I = 17.99 A

I = 18 A

Therefore, the current will be 18 A

Answer:

The energy density is  [tex]Z = 5.4 2 *10^{-5 } \ J/m^3[/tex]

Explanation:

From the question we are told that

    The electric vector is  [tex]E = 3500 \ V/m[/tex]

Generally the energy vector is mathematically represented as

      [tex]Z = 0.5 * \epsilon_o * E^2[/tex]

Where  [tex]\epsilon_o[/tex] is the permitivity of free space with the value  [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

      [tex]Z = 0.5 * 8.85 *10^{-12} * 3500^2[/tex]

      [tex]Z = 5.4 2 *10^{-5 } \ J/m^3[/tex]

Answer:

d. four times as far

Explanation:

Initial velocity of car and truck, u = 0

let acceleration of both the truck and car = a

let the length of time for the acceleration = t

Let the time the truck accelerated = 2t

The distance traveled by the car is calculated as;

s = ut + ¹/₂at²

s₁ = 0(t) + ¹/₂at²

s₁ = ¹/₂at²

The distance traveled by the truck is calculated as;

s = ut + ¹/₂at²

s₂ = 0(2t) + ¹/₂a (2t)²

s₂ =  ¹/₂a x 4t²

s₂ = 4 (¹/₂at²)

s₂ = 4(s₁)

Truck distance = four times car distance

Therefore, Compared with the car, the truck will travel four times as far

d. four times as far

Answer:

3.21×10⁻³ Ωm

Explanation:

Applying,

R = Lρ/A................... Equation 1

Where R = Resistance of the wire, L = Length of the wire, ρ = Resistivity of the wire, A = cross sectional area of the wire.

Make ρ the subject of the equation

ρ = RA/L................... Equation 2

Given: R = 0.31 Ohms, A = 0.003 m², L = 0.29 m

Substitute into equation 2.

ρ = 0.31(0.003)/0.29

ρ  = 3.21×10⁻³ Ωm

Answer:

Explanation:

Load will be moved by 4L when effort moves by distance L .

4L = 10 m ( given )

L = 2.5 m

work output = work input = 400 x 10 = 4000 J

work by friction = 100 J

net work output = 3900 J .

efficiency = net output of work / work input

= (3900 / 4000) x 100

= 97.5 %

2 )

work input = 4000 J

distance moved by effort = 2.5 m

If effort be F

F X 2.5  = 4000

F = 1600 N .

Answer:

The  initial velocity is  [tex]v_h = 8.66 \ m/s[/tex]

Explanation:

From the question we are told that

    The height of the tree is  [tex]h = 3.30\ m[/tex]

    The distance of the position of landing from base  is  [tex]d = 5.30 \ m[/tex]

According to the second equation of motion

    [tex]h = u_o * t + \frac{1}{2} at^2[/tex]

[tex]Where\ u_o[/tex] is the initial velocity in the vertical axis  

           a  is equivalent to acceleration due to gravity which is positive because the tiger is downward

    So

     [tex]3 = 0 + 0.5 * 9.8 *t^2[/tex]

=>    [tex]t = \frac{3 }{9.8 * 0.5}[/tex]

      [tex]t = 0.6122\ s[/tex]

Now the initial velocity in the horizontal direction is mathematically evaluated as

         [tex]v_h = \frac{5.30}{0.6122}[/tex]

        [tex]v_h = 8.66 \ m/s[/tex]

Answer:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

See attached file pls

Answer:

Explanation:

For current in ideal transformer the formula is

I₁ / I₂ = N₂ / N₁

I₁  and I₂ are current in primary and secondary coil respectively and N₁ and N₂ are no of turns in primary and secondary coil .

Putting the given values

100 / I₂ = 200 / 100 = 2

I₂ = 50 A .

output current = 50 A .

Answer:

the water that remain in the tank in one hour will be 200 lb

Explanation:

Initial mass of water in the tank = 2000 lb

hot water is delivered through the first inlet pipe at a rate of = 0.8 lb/s

cold water is delivered through the second inlet pipe at a rate of = 1.2 lb/s

exit pipe flow rate = 2.5 lb/s

amount of water in the tank after one hour = ?

In one hour, there are 60 x 60 seconds = 3600 sec, therefore

the water through the first inlet pipe in one hour = 0.8 x 3600 = 2880 lb

the water through the second inlet pipe in one hour = 1.2 x 3600 = 4320 lb

the water through the exit in one hour = 2.5 x 3600 = 9000 lb

The total amount of water in the tank = 2000 + 2880 + 4320 = 9200 lb

The total amount of water that leaves the tank = 9000 lb

therefore, in one hour, the water that remain in the tank will be

==> 9200 lb - 9000 lb = 200 lb

Answer:

The magnitude of flux remains the same, and the field increases.

Explanation:

This is because the number of field lines leaving the sphere remains constant and the electric field increases because the line density increases

Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet [tex]v_{1}[/tex] = 250 m/s

final velocity of bullet [tex]v_{2}[/tex] = 140 m/s

loss of kinetic energy of the bullet = [tex]\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})[/tex]

==> [tex]\frac{1}{2}*0.0033*(250^{2} - 140^{2} )[/tex] = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = [tex]\frac{1}{2} mv^{2}[/tex]

70.785 = [tex]\frac{1}{2}*0.25*v^{2}[/tex]

566.28 = [tex]v^{2}[/tex]

[tex]v= \sqrt{566.28}[/tex] = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s